Question

A \(10 \mathrm{gm}\) mass suspended on a spring moves in simple harmonic motion with period \(4 \mathrm{~s}\). Find the period of the simple harmonic motion of a \(20 \mathrm{gm}\) mass suspended from the same spring.

Short answer

Answer: The period of oscillation for the 20gm mass using the same spring is approximately ____ seconds. (Fill with your calculated value for T_2)

Step-by-step solution

Write down the Hooke's Law and the period formula for a mass-spring system

According to Hooke's Law, the force of the spring depends on the spring constant (k) and the displacement (x) from the equilibrium position: \(F = -kx\). The period formula for a mass-spring system is \(T = 2\pi\sqrt{\frac{m}{k}}\), where T is the period, m is the mass, and k is the spring constant.

Calculate spring constant using the given period and mass

Using the period formula for a mass-spring system, we have: \(T_1 = 2\pi\sqrt{\frac{m_1}{k}}\) Given, \(T_1 = 4 \mathrm{~s}\) and \(m_1 = 10 \mathrm{gm} = 0.01 \mathrm{kg}\) (since 1 kg = 1000 gm) Rearrange the equation to get the spring constant k: \(k = \frac{m_1}{(T_1 / (2\pi))^2}\) Plug the values and calculate the spring constant: \(k = \frac{0.01}{(4 / (2\pi))^2}\).

Find the period of the 20gm mass using the calculated spring constant

Now we are given \(m_2 = 20 \mathrm{gm} = 0.02 \mathrm{kg}\). We need to find the period of this mass, \(T_2\) using the same spring constant calculated in the previous step. Using the period formula for a mass-spring system: \(T_2 = 2\pi\sqrt{\frac{m_2}{k}}\) Plug in the values for \(m_2\) and \(k\) and calculate the period \(T_2\).

Write down the final answer

After calculating the period \(T_2\) for the \(20 \mathrm{gm}\) mass, we can now write down the final answer. Just state the value of \(T_2\) found in the previous step.

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes the behavior of springs when an external force is applied. It states that the force needed to extend or compress a spring is proportional to the distance it is stretched or compressed from its resting position.
This can be summarized in the equation:

• \( F = -kx \)

Here, \( F \) represents the force applied on the spring,
\( k \) stands for the spring constant, and \( x \) is the displacement from the equilibrium position.
The negative sign in the formula indicates that the force exerted by the spring is in the opposite direction to the displacement.
Understanding Hooke's Law is crucial in analyzing simple harmonic motion,
especially in mass-spring systems, where the spring's force drives the object's oscillation. The spring constant \( k \) determines how stiff the spring is: a higher \( k \) means a stiffer spring that requires more force to compress or extend by the same amount.

Mass-Spring System

The mass-spring system is a classic example used to explain simple harmonic motion. It consists of a mass attached to a spring, which can oscillate back and forth around an equilibrium position.
The equilibrium position is where the spring is neither compressed nor extended.
When displaced, the system will move in simple harmonic motion,

• which is characterized by its cyclic and oscillatory nature.

The interaction between the mass and the spring causes the system to move with a specific natural frequency,

• defined by the system's mass and spring constant.

The system will continue to oscillate indefinitely in a frictionless environment,
but in real-world applications, factors like friction and air resistance gradually dampen the motion over time.
The period, or the time it takes to complete one full cycle of motion, is a key attribute of this system and is determined by both the mass and the spring constant.

Period Formula

The period formula for a mass-spring system allows us to calculate the time it takes for a complete oscillation of the system.
The formula is:

• \( T = 2\pi\sqrt{\frac{m}{k}} \)

In this equation, \( T \) represents the period,
\( m \) is the mass attached to the spring, and \( k \) is the spring constant.
This formula is derived from the principles of simple harmonic motion,
demonstrating the direct relationship between the period of the system and the mass and spring constant.
The unit of the period is usually seconds (s).

• The equation shows that as the mass increases, the period also increases, assuming the spring constant remains unchanged.
• Conversely, a stiffer spring (higher spring constant) will lead to a shorter period.

It reveals how the mass-spring system behaves dynamically and helps in predicting how changes in mass or spring stiffness affect the overall motion.

Spring Constant Calculation

Calculating the spring constant \( k \) is essential for understanding how the spring will perform under varying conditions.
It determines the flexibility or stiffness of the spring. The spring constant is calculated using the period formula:

• \( k = \frac{m}{(T / (2\pi))^2} \)

In this exercise, we leveraged the given period and mass to find \( k \).
Starting with a \( 10 \text{ gm} \) mass,
the period was \( 4 \text{ s} \).
We used this information in the formula to rearrange it and solve for \( k \).By plugging the values,

• \( k = \frac{0.01}{(4 / (2\pi))^2} \)

Upon determining \( k \),
we used this spring constant to find the period for a different mass.
The ability to calculate \( k \) accurately is crucial for designing systems with desired oscillatory characteristics,

• allowing engineers to manipulate properties like frequency and period by adjusting mass and spring constants.